What does ${3#?} do?

That is called parameter expansion:

1. $3 is your variable, which is the third parameter of the script/function.
2. # will remove the shortest prefix of the variable.
3. ? is the pattern you are looking for (in this case is any character).

So basically you remove the shortest prefix of the variable named 3 until you find the pattern.

In general, we can consider:

${variable_name[option][pattern]} 

There are other options like:

• ## remove largest prefix.
• % remove shortest suffix.
• %% remove largest suffix.

Which you can combine with other patterns, for example, for getting the last field of a CSV line:

> string="asdf,1234,aa,foo22" > echo ${string##*,} foo22 

Notice how we have removed the largest prefix searching for the pattern "any character(s) followed by a comma".

> set -- 1 2 foo > echo "$3" foo TEST_VARIABLE=${3#?} > echo "$TEST_VARIABLE" oo 

It assigns the value of the third positional parameter without its first char to the variable TEST_VARIABLE (the positional parameter itself is not changed); used in functions or shell scripts:

./myscript 1 2 foo # or myfunc 1 2 foo # within each $3 is foo 

In order to have positional parameters within an interactive shell you need set.